જો (sin ^{-1} x)^{2}-(cos ^{-1} x)^{2}=a ; 0 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) આપેલ છે કે $a = (\sin ^{-1} x)^{2}-(\cos ^{-1} x)^{2}$.નિત્યસમ $A^2 - B^2 = (A+B)(A-B)$ નો ઉપયોગ કરતા:$a = (\sin ^{-1} x + \cos ^{-1} x)(\sin ^{-1

Vedclass · Accepted Answer

$\sin \left(\frac{2 a}{\pi}\right)$

Vedclass · Answer

(B) આપેલ છે કે $a = (\sin ^{-1} x)^{2}-(\cos ^{-1} x)^{2}$.નિત્યસમ $A^2 - B^2 = (A+B)(A-B)$ નો ઉપયોગ કરતા:$a = (\sin ^{-1} x + \cos ^{-1} x)(\sin ^{-1} x - \cos ^{-1} x)$.આપણે જાણીએ છીએ કે $\sin ^{-1} x + \cos ^{-1} x = \frac{\pi}{2}$,તેથી:$a = \frac{\pi}{2}(\sin ^{-1} x - \cos ^{-1} x)$.$\sin ^{-1} x = \frac{\pi}{2} - \cos ^{-1} x$ મૂકતા:$a = \frac{\pi}{2}(\frac{\pi}{2} - 2 \cos ^{-1} x)$.$\frac{2a}{\pi} = \frac{\pi}{2} - 2 \cos ^{-1} x$.$2 \cos ^{-1} x = \frac{\pi}{2} - \frac{2a}{\pi}$.બંને બાજુ કોસાઇન (cosine) લેતા:$\cos(2 \cos ^{-1} x) = \cos(\frac{\pi}{2} - \frac{2a}{\pi})$.નિત્યસમ $\cos(2 	heta) = 2 \cos^2 	heta - 1$ અને $\cos(\frac{\pi}{2} - 	heta) = \sin 	heta$ નો ઉપયોગ કરતા:$2 x^2 - 1 = \sin(\frac{2a}{\pi})$.

જો $(\sin ^{-1} x)^{2}-(\cos ^{-1} x)^{2}=a ; 0 < x < 1, a \neq 0$ હોય,તો $2 x^{2}-1$ ની કિંમત શું થાય?

Similar Questions

કિંમત શોધો: $\operatorname{cosec}^{-1}\left[\left(\frac{\tan ^2\left(\frac{\alpha-\pi}{4}\right)-1}{\tan ^2\left(\frac{\alpha-\pi}{4}\right)+1}+\cos \frac{\alpha}{2} \cdot \cot 5 \alpha\right) \sec \frac{11 \alpha}{2}\right]$

$\cos ^{-1}\left[\frac{1}{\sqrt{2}}\left(\cos \frac{9 \pi}{10}-\sin \frac{9 \pi}{10}\right)\right]$ નું મુખ્ય મૂલ્ય શોધો.

જો $\cos^{-1}\left(\frac{x}{a}\right) + \cos^{-1}\left(\frac{y}{b}\right) = \alpha$ હોય,તો $\frac{x^2}{a^2} - \frac{2xy}{ab}\cos \alpha + \frac{y^2}{b^2} = $

$2 \tan ^{-1}\left(\frac{1}{3}\right)+\cos ^{-1}\left(\frac{3}{5}\right)=$

જો $\tan ^{-1}(2 x)+\tan ^{-1}(3 x)=\frac{\pi}{4}$,જ્યાં $x>0$,તો $x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module